zoukankan      html  css  js  c++  java
  • HDU

    E - Super Jumping! Jumping! Jumping!

     

    Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. 



    The game can be played by two or more than two players. It consists of a chessboard(棋盘)and some chessmen(棋子), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path. 
    Your task is to output the maximum value according to the given chessmen list. 
    Input
    Input contains multiple test cases. Each test case is described in a line as follow:
    N value_1 value_2 …value_N 
    It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int. 
    A test case starting with 0 terminates the input and this test case is not to be processed. 
    Output
    For each case, print the maximum according to rules, and one line one case. 
    Sample Input
    3 1 3 2
    4 1 2 3 4
    4 3 3 2 1
    0
    Sample Output
    4
    10
    3


    确定状态:dp[i]表示以i数字为结尾的单调递增子序列

    状态转移方程:dp[i]=max(dp[i],dp[j]+arr[i]);

    求单调递增子序列  

    #include <bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define maxn 1010
    
    int arr[maxn];
    int dp[maxn];///dp[i]是以第i个字母结尾的递增子序列的个数
    
    int main() {
        int n;
    	while(~scanf("%d",&n) && n!=0)
    	{
    	    for(int i=0;i<n;i++)
    		    scanf("%d",&arr[i]);
    		int ans=0;
    		for(int i=0;i<n;i++)
    		{
    			dp[i]=arr[i];
    			for(int j=0;j<i;j++)
                    if(arr[i]>arr[j])
    				  dp[i]=max(dp[i],dp[j]+arr[i]);
    		}
    		int mmax=-INF;
    		for(int i=0;i<n;i++)
                mmax=max(mmax,dp[i]);
    		cout<<mmax<<endl;
    	}
    	return 0;
    }
    




  • 相关阅读:
    day40_jQuery学习笔记_01
    jQuery选择什么版本 1.x? 2.x? 3.x?
    6个关于dd命令备份Linux系统的例子
    快速掌握grep命令及正则表达式
    Linux下删除乱码或特殊字符文件
    在 Linux 中永久修改 USB 设备权限
    CentOS 7 中 hostnamectl 的使用
    申请红帽企业版Linux开发者订阅
    CentOS6 下rsync服务器配置
    Centos6下DRBD的安装配置
  • 原文地址:https://www.cnblogs.com/zswbky/p/6792920.html
Copyright © 2011-2022 走看看